import java.awt.event.AdjustmentEvent;

/**
 * @author : WXY
 * @create : 2022-09-04 21:10
 * @Info : 斐波那契数列
 */
public class Code01_FibonacciProblem {
    //f函数 正常的斐波那契数列问题
    public static int f1(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2) {
            return 1;
        }
        return f1(n - 1) + f1(n - 2);
    }

    public static int f2(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2) {
            return 1;
        }
        int res = 1;
        int pre = 1;
        int tmp = 0;
        for (int i = 3; i <= n; i++) {
            tmp = res;
            res = res + pre;
            pre = tmp;
        }
        return res;
    }

    public static int f3(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2) {
            return 1;
        }
        // [ 1 ,1 ]
        // [ 1, 0 ]
        int[][] base = {
                {1, 1},
                {1, 0}
        };
        int[][] res = matrixPower(base, n - 2);
        return res[0][0] + res[1][0];
    }

    public static int[][] matrixPower(int[][] m, int p) {
        int[][] res = new int[m.length][m[0].length];
        for (int i = 0; i < res.length; i++) {
            res[i][i] = 1;
        }

        // res = 矩阵中的1
        int[][] tmp = m;// 矩阵1次方
        for (; p != 0; p >>= 1) {
            if ((p & 1) != 0) {
                res = muliMatrix(res, tmp);
            }
            tmp = muliMatrix(tmp, tmp);
        }
        return res;
    }

    // 两个矩阵乘完之后的结果返回
    public static int[][] muliMatrix(int[][] m1, int[][] m2) {
        int[][] res = new int[m1.length][m2[0].length];
        for (int i = 0; i < m1.length; i++) {
            for (int j = 0; j < m2[0].length; j++) {
                for (int k = 0; k < m2.length; k++) {
                    res[i][j] += m1[i][k] * m2[k][j];
                }
            }
        }
        return res;
    }

    // s函数一个小人上楼梯问题，只能向上登1步或者2步 ,登到n 台阶，一共多少中方法
    public static int s1(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2) {
            return n;
        }
        return s1(n - 1) + s1(n - 2);
    }

    public static int s2(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2) {
            return n;
        }
        int res = 2;
        int pre = 1;
        int tmp = 0;
        for (int i = 3; i <= n; i++) {
            tmp = res;
            res = res + pre;
            pre = tmp;
        }
        return res;
    }

    public static int s3(int n) {
        if (n < 1) {
            return  0;
        }
        if (n == 2 || n == 1) {
            return n;
        }
        int[][] base = {{1, 1}, {1, 0}};
        int[][] res = matrixPower(base, n - 2);
        return 2 * res[0][0] + res[1][0];
    }


    //c函数是奶牛问题，
    //一头奶牛，永远都不会死，每一个新出生的奶牛在出生后的第三年开始生新的一头奶牛
    //
    public static int c1(int n) {
        if (n < 1) {
            return  0;
        }
        if (n == 1 || n == 2 || n == 3) {
            return n;
        }
        return c1(n - 1) + c1(n - 3);
    }

    //这个常数级别的函数c2没有仔细看，
    public static int c2(int n) {
        if (n < 1) {
            return  0;
        }
        if (n == 1 || n == 2 || n == 3) {
            return n;
        }
        int res = 3;
        int pre = 2;
        int prepre = 1;
        int tmp1 = 0;
        int tmp2 = 0;
        for (int i = 4; i <= n; i++) {
            tmp1 = res;
            tmp2 = pre;
            res = res + prepre;
            pre = tmp1;
            prepre = tmp2;
        }
        return res;
    }

    public static int c3(int n) {
        if (n < 1) {
            return 0;
        }
        if (n == 1 || n == 2 || n == 3) {
            return n;
        }
        int[][] base = {
                {1, 1, 0},
                {0, 0, 1},
                {1, 0, 0}};
        int[][] res = matrixPower(base, n - 3);
        return 3 * res[0][0] + 2 * res[2][1] + 1 * res[2][0];
    }

    public static void main(String[] args) {
        int n = 19;
        System.out.println(f1(n));
        System.out.println(f2(n));
        System.out.println(f3(n));
        System.out.println("======================");

        System.out.println(s1(n));
        System.out.println(s2(n));
        System.out.println(s3(n));
        System.out.println("======================");

        System.out.println(c1(n));
        System.out.println(c2(n));
        System.out.println(c3(n));

    }
}
